POJ-2387 Til the Cows Come Home ( 最短路 )

POJ 2387 最短路径问题

最新推荐文章于 2020-07-10 23:31:10 发布

转载最新推荐文章于 2020-07-10 23:31:10 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/hollowstory/p/5547105.html

本文介绍了一道经典的最短路径问题，题目要求从N个地标中找到从N号地标回到1号地标（即谷仓）的最短路径。使用双向路径和特定算法解决该问题，并给出了完整的C++实现代码。

题目链接: http://poj.org/problem?id=2387

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

Sample Output

90

最短路模板题，求1到n的最短路

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<stack>
 6 #include<queue>
 7 
 8 using namespace std;
 9 
10 int way[1005][1005];
11 bool flag[1005];
12 int main(){
13     ios::sync_with_stdio( false );
14 
15     int n, m;
16     
17     while( cin >> m >> n ){
18         int x, y, d;
19         memset( way, 0x3f3f3f3f, sizeof( way ) );
20         memset( flag, false, sizeof( flag ) );
21         for( int i = 0; i < m; i++ ){
22             cin >> x >> y >> d;
23             way[x][y] = way[y][x] = min( way[x][y], d );
24         }
25 
26         for( int k = 0; k < n - 1; k++ ){
27             int minv = 0x3f3f3f3f, mini;
28 
29             for( int i = 1; i < n; i++ ){
30                 if( !flag[i] && minv > way[n][i] ){
31                     minv = way[n][i];
32                     mini = i;
33                 }
34             }
35 
36             flag[mini] = true;
37             for( int i = 1; i < n; i++ ){
38                 if( !flag[i] ){
39                     way[n][i] = min( way[n][i], way[n][mini] + way[mini][i] );
40                 }
41             }
42         }
43 
44         cout << way[n][1] << endl;
45     }
46 
47     return 0;
48 }

转载于:https://www.cnblogs.com/hollowstory/p/5547105.html